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### Network # Deciding factor with potential dividers

I

#### Ian Tedridge

Jan 1, 1970
0

If I want to produce a potential divider with a 5Vdc supply. Wanting 2Vdc
out. I guess i would use 2 equal rated resistors in series.

How do I decide what size resistors 10K or 20K etc ?

I dont seem to understand the importance of resistor size.

F

#### Fred Stevens

Jan 1, 1970
0
Ian said:

If I want to produce a potential divider with a 5Vdc supply. Wanting 2Vdc
out. I guess i would use 2 equal rated resistors in series.

How do I decide what size resistors 10K or 20K etc ?

I dont seem to understand the importance of resistor size.

How much current does your load need to draw at 2V?

fred

R

#### Ralph Mowery

Jan 1, 1970
0
Ian Tedridge said:

If I want to produce a potential divider with a 5Vdc supply. Wanting 2Vdc
out. I guess i would use 2 equal rated resistors in series.

How do I decide what size resistors 10K or 20K etc ?

I dont seem to understand the importance of resistor size.

You just can not say you want 2 volts out. You must specify the load
resistance. If the 2 volt load is very high in resistance you usually will
use a high value of resistance. If the load is low in resistance , you will
have to use low values of resistance. It will also depend on some extent of
the source to supply the needed current. If for example you want to get
2.5 volts from a simple two resistor devider into a high resistance load and
you use a pair of 5 ohm reisitors in series across the load , the source
must be able to supply 1/2 of an amp. This is usually more wasted power
than you want. If you keep the resistors so they are more than 100 times
the load resistance the error is usually small enough not to be noticed.
As with most electronic circuits , there is a trade off in the components
used with the power and price of them.

I

#### Ian Tedridge

Jan 1, 1970
0
Looking to pull 250mA Max

R

#### Ralph Mowery

Jan 1, 1970
0
Ian Tedridge said:
Looking to pull 250mA Max
To reduce a 5 volt source to a 250 ma load you should not be looking at
resistors but another way. Maybe 6 silicon 1 amp diodes in series with the
load. The reisitors will not hold the voltage constant under a changing

D

#### David L. Jones

Jan 1, 1970
0
Ian said:
Looking to pull 250mA Max

You won't really be able to use a resistor divider to provide 250mA,
that is *way* too much current. You are better off using say a LM317
voltage regulator to drop the 5V input to give you 2V output. Then it
will be 2V output regardless of the load you put on. A resistor divider
voltage output will change with the current your load draws.

Dave J

#### Jonathan Kirwan

Jan 1, 1970
0
If I want to produce a potential divider with a 5Vdc supply. Wanting 2Vdc
out. I guess i would use 2 equal rated resistors in series.

If by "equal rated" you mean equal valued, as in ohms, no. That would
provide 2.5V, not 2V.
How do I decide what size resistors 10K or 20K etc ?

I dont seem to understand the importance of resistor size.

I'm assuming you imagine this:

: +5V
: ,----------,
: | |
: | |
: | \
: | / R1
: | \ ???
: | /
: | |
: --- | +2V
: - V1 +--------,
: --- 5V | |
: - | |
: | \ \
: | / R2 / Rload
: | \ ??? \ 2V @ 250mA
: | / /
: | | |
: | | |
: '----------+--------'

In this case, your Rload resistance is as little as 2V/.25 or 8 ohms.
But let's assume that it can draw a lot less current (for example, 1mA
would suggest 2000 ohms, 1uA would suggest 2M ohm, etc.) So one of
the first key questions to ask yourself is:

What is the range of voltage I want to accept at the +2V node?

I think you can see that if the answer is 2.0000 to 2.0000, it's going
to be practically impossible to achieve. No tolerance of a voltage
variation is impossible, if the load will itself vary it's loading.
You have to have some finite, non-zero range you accept.

So let's say it is from 2.2V to 1.9V, with 2.2V being the voltage when
Rload isn't even connected (no loading) and 1.9V being the case when
Rload requires 250mA. First off, with no load, you just have R1 and
R2 and they must develop 2.2V: V1*R2/(R1+R2) = 2.2V, V1 = 5V.

With Rload drawing current of 250mA, it's easier to imagine that the
circuit has been transformed into this Thevenin equivalent:

: +2.2V
: ,----------,
: | |
: | |
: | \
: | / Rth
: | \ ???
: | /
: | |
: --- |
: - Vth +-- +1.9V
: --- |
: - |
: | \
: | / Rload
: | \ 1.9V @ 250mA
: | /
: | |
: | |
: '----------'

I'm not sure if you have already read about Thevenin, but the basic
idea is that the more complex combination of V1, R1 and R2 can be
replaced by a new, simpler Vth and Rth, which is what your Rload
"sees" looking back into the V1, R1, and R2 circuit. If you haven't,
this all may seem a bit strange and I'd recommend doing some reading
and thinking on the idea on your own. It's not hard to follow, but I
don't want to side-track the discussion right now with it. So just
accept the above equivalent.

Now Rth is just R1 in parallel to R2, so we have Rth = R1*R2/(R1+R2).
The difference in the 1.9V node and the 2.2V value for Vth is 0.3V.
And we already know what current Rload is drawing in this case, 250mA.
So that means Rth must be .3V/.25A or 1.2 ohms total. So we then know
that: R1*R2/(R1+R2) = 1.2 ohms.

This gives us:

(1) 5*R2/(R1+R2) = 2.2 --or-- R2/(R1+R2) = 2.2/5
--and also that--
(2) R1*R2/(R1+R2) = 1.2

So, substituting (1) into (2), we get:

(2) R1*2.2/5 = 1.2

Which works out as R1 = 1.2 * 5 / 2.2, or 30/11 or about 2.7 ohms.

R2 is then simply R2/(30/11+R2) = 2.2/5, or R1*(2.2/5)/(1-2.2/5),
which works out to 15/7 or about 2.1 ohms.

This would mean that your divider would be normally consuming about 1A
of current. In other words, about 5 watts of power would be
dissipated in R1 and R2, without ever hooking up your load. With the
load in place, more than 5.5 watts total.

This is usually impractical, which is why other suggestions were
offered.

Jon

R

#### redbelly

Jan 1, 1970
0
To put it another way, and more briefly:

Suppose you want to regulate the voltage to within, say, 10% of the 2V.
That means choosing divider resistors that will draw TEN TIMES the max
load current: 10 x 0.25A or 2.5A.

This means using a power supply that is considerably costlier than what
you'd need by using a different method.

Best to use a voltage regulator, so that the power source need only
supply slighly more current than the load's 250 mA. AND the regulation
will be way better than 10%!.

Regards,

Mark

J

#### Jonathan Kirwan

Jan 1, 1970
0
: +5V
: ,----------,
: | |
: | |
: | \
: | / R1
: | \ ???
: | /
: | |
: --- | +2V
: - V1 +--------,
: --- 5V | |
: - | |
: | \ \
: | / R2 / Rload
: | \ ??? \ 2V @ 250mA
: | / /
: | | |
: | | |
: '----------+--------'

Suppose you transform this, using an NPN transistor:

: +5V
: ,----------+-------,
: | | |
: | | | almost 250mA
: | \ | |
: | / R1 | |
: | \ ??? | v
: | / |
: | | |
: --- | |/ Q1
: - V1 +-----| NPN
: --- 5V | |>-,
: - | | <---- supposed to be about 2V
: | \ \
: | / R2 / Rload
: | \ ??? \ 2V @ 250mA
: | / /
: | | |
: | | |
: '----------+--------'

Now, in this case you can use far less current in your R1,R2 divider
because the transistor's base will not divert much current away from
it.

Q1 will bring pretty much all of the 250mA used in Rload via the
collector that connects directly back to the +5 side of the 5V battery
or power supply. As a 0-order estimate of Q1's base voltage, you can
just assume that the base-to-emitter voltage (Vbe) is somewhere
between 0.6V (lower Vbe voltages occur with smaller base currents,
larger Vbe with larger base currents, and it is unusual for 0.6V to be
the case unless we are talking about fairly low currents overall) to
about 0.9V or slightly more (for times when the transistor is involved
with fairly high currents.)

Normally, this variation of Vbe is about 60mV for each factor of 10
change in current, so you can see that it varies only slightly. So
let's say it runs at about 750mV with a base current of 100uA, just to
pick some figure to start. If the base current went to 1mA, you'd
expect the Vbe voltage to then be about 750mV + 60mV or about 810mV.
That should give you a rough picture, here.

Getting back to the diagram... Rload needs about 250mA per your
specification. This means a collector current of about the same
value. But what base current? (We need to have a rough idea of the
base current to estimate Vbe -- not that it is terribly important to
do, as it isn't because of the log() behavior, but let's do it
anyway.)

Well, if the transistor is to be operated in saturation that would be
one thing. But it isn't in this case. The collector voltage is known
to be 5V and the emitter is at 2V, roughly. What causes a transistor
to go into saturation, so to speak, is when the voltage potential at
the collector is such that it begins to even slightly forward bias the
base to collector diode in it (it's supposed to be normally reverse
biased.) In this case, we already know that the base won't be more
than about 3V (we are still trying to figure exactly what, but we can
delay this for now) and with the collector at 5V the internal base to
collector diode will definitely be reverse biased. So no saturation

Okay, so that means that we can anticipate a transistor beta of say
100 or more. Very old transistors, like point-contact types, might
have betas in the 25-30 range. But none of those I've used recently
are that "bad" unless they are in reasonable saturation (when the
forward biased base-to-collector diode demolishes any decent beta
computation.) You could probably even be safe thinking 200. But
let's call it 100 to be extra safe in the other figures we derive.

So the base current will be about 1/100th of the 250mA. This means...
2.5mA. Going back to the 100uA starting point I mentioned, this is a
factor of 25 times more or about a fourth of 10x10. The first 10x
gets us another 60mV. Let's not get into log() functions and just
accept that the next 2.5 times will about to 1/4th of the 60mV more,
or 15mV. So that means 75mV or so above our guess of 750mV for the
NPN, or 825mV. Call it 850mV.

Okay, so you want the base to be about 850mV above 2.0V or 2.85V. But
notice now that your load on the R1,R2 divider has been reduced to the
base current which we estimate at 2.5mA -- maybe even less than that.
This means that all of the same calculations performed before apply
here again -- except that you now what the divider to provide 2.85V
instead of the original 2V or so and that you now only need to supply
2.5mA or so instead of 250mA.

With no load at all, or better for thinking about, close to no load
but with a very slight load, the Vbe will be around 0.6V or so. This
means that if we want there to be 2.2V unloaded and 1.9V loaded down,
we want the base to be 2.2+0.6 or 2.8V unloaded and then about 1.9+.85
or 2.75V loaded. A change of 50mV total. Remember this figure, as
I'll use it shortly in (2) below.

Referring back to the old equations and the Thevenin equivalent,
updated now with new values, we have:

(1) 5*R2/(R1+R2) = 2.8 --or-- R2/(R1+R2) = 2.8/5
--and also that--
(2) Rth=R1*R2/(R1+R2) = dV/dI = (50mV/2.5mA) = 20

The 50mV comes from the voltage change noted above. The 2.5mA comes
from the change in base current between an unloaded case and a loaded
one.

This solves out to R1 = 20*5/2.8 = 100/2.8 = 35.7 ohms and to R2 =
R1*(2.8/5)/(1-2.8/5) = 45 4/9 = 45.5 ohms.

Now we are talking about only 60mA or so through the divider with
250mA through Rload and the wattage in the divider is about 300mW

There will be another (5V - 1.9V)*250mA or 775mW in the transistor, as
well. But that waste is unavoidable in linear (non-switching) designs
because you simply _have_ to throw away about 3V at 250mA no matter
what you do. So that 3/4 watt is going to be wasted somewhere.

So this design is about 300mW more than the theoretical best case for
a linear arrangement (which must be 750mW, as I argue) rather than
being more than 4W beyond, like the simple divider was. And all you
had to do was add a transistor to get there and redo a few
calculations for the resistors.

You could plan on further reductions on the base current by adding
another transistor. But all you could hope to do is eat into that
300mW and reduce it somewhat. But that 750mW waste will always remain
no matter how many transistors you stack up, unless you consider a
switcher arrangement.

Jon

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